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Posts Tagged ‘regular local rings’

## Regular implies locally factorial

I’ve decided to end the series of posts on geometric class field theory (not that I’d been posting much anyway). This is partially because I’m in the process of writing up my own interpretation and proofs of these results, which will hopefully appear eventually. The other reason is that Deligne’s proof, which is the one I was going to present, is explained well in several places, e.g. Edward Frenkel’s article “Lectures on the Langlands program and conformal field theory” and Peter Toth’s master’s thesis.

Instead, I’ll try to post here when I encounter ideas which I think aren’t well enough documented and deserve to be written up. Today I want to prove the often stated but rarely proved fact that a regular local ring is a unique factorization domain. According to Wikipedia, in 1958 Nagata reduced the theorem to the case of a three-dimensional regular local ring, and in 1959 Auslander and Buchsbaum proved the result in dimension three. Their proof has been simplified over the years, although it still involves a fairly complicated induction on dimension, and in my mind doesn’t give much insight into why the theorem is true. Below I’ll present a beautiful geometric argument which was explained to me by Dennis Gaitsgory. Much of what follows is based on unpublished notes from a class he taught on the theory of schemes.

Although we are trying to prove a theorem in commutative algebra, we will use non-affine schemes in the proof. Let’s tautologically reformulate the statement in scheme theory.

Recall that a scheme is called regular provided that all its local rings are regular. Similarly, a scheme is called locally factorial provided that all its local rings are unique factorization domains, hereafter abbreviated to UFDs. So we are trying to prove the following.

Theorem A regular scheme is locally factorial.

There is some potential for confusion in the term “locally factorial.” If a scheme $X$ has an open covering by affine schemes $U_i = \text{Spec } A_i$ with each $A_i$ a UFD, then $X$ is factorial, but the converse is false. In fact, it may happen that a factorial scheme contains no affine open subschemes $U = \text{Spec } A$ with $A$ a UFD, e.g. any smooth curve of positive genus over a field.

We’ll use the following geometric criterion for local factoriality. Recall that $X$ is normal provided that the local ring at every point is an integrally closed domain. With everything in sight being a property of local rings, it suffices to formulate the criterion for affine schemes. Actually, we will assume throughout that $X = \text{Spec } A$ is noetherian and integral as well, i.e. that $A$ is a noetherian domain. The noetherian hypothesis is essential for the kind of results we’re considering, but integrality just makes for cleaner statements: if the local rings of $X$ are domains then $X$ is the disjoint union of its irreducible components (because $X$ is noetherian!), which are themselves integral. So we are merely avoiding connectedness hypotheses and related complications.

Proposition 1 A noetherian affine integral scheme $X$ is locally factorial if and only if

1. $X$ is normal, and
2. for any open $U \subset X$ whose complement has codimension at least two, every line bundle on $U$ extends to a line bundle on $X$.

Thus an arbitrary locally noetherian scheme is locally factorial if and only if it can be covered by open affines satisfying the criterion above.

If this is what local factoriality really means, it begs the question: what does normality really mean? We’ll need to answer this question before we prove Proposition 1. Recall that a scheme is regular in codimension one provided that all of its one-dimensional local rings are regular. This hypothesis helps us to pass from Cartier divisors to Weil divisors, since it ensures that the local ring at a codimension one point is a discrete valuation ring.

Proposition 2 A noetherian integral affine scheme $X$ is normal if and only if $X$ is regular in codimension one and the following equivalent conditions hold:

1. no effective Cartier divisor on $X$ has an embedded point,
2. for some (equivalently, any) line bundle $\mathscr{L}$ and $j : U \to X$ the inclusion of an open set whose complement has codimension at least two, the canonical morphism $\mathscr{L} \to j_*j^*\mathscr{L}$ is an isomorphism, and
3. Serre’s condition $\text{S}_2$ holds, i.e. for any coherent sheaf $\mathscr{F}$ on $X$ whose support has codimension at least two, we have $\text{Ext}^1(\mathscr{F},\mathscr{O}_X) = 0$.

Condition 1 is pleasantly geometrical while condition 3 is just technical, but condition 2 deserves further discussion. It might be called the “Hartogs condition,” after the Hartogs theorem in complex analysis which says that a holomorphic function extends over loci of codimension at least two. We might also compare it to condition 2 in Proposition 1, which says that line bundles extend over loci of codimension at least two, whereas condition 2 in Proposition 2 says that such an extension, if it exists, is unique.

We will prove the statements above in the order opposite to the one in which they were stated, but first we need a lemma.

Lemma 1 Let $X$ be a noetherian integral affine scheme. Then a rational function on $X$ is regular if and only if it is regular along every associated point of every effective Cartier divisor.

Proof. For the non-tautological direction, suppose we are given a rational function $\tfrac{f}{g}$ on $X = \text{Spec } A$ which is not regular, so $f \notin gA$. Then $I = (fA+gA)/gA \neq 0$: geometrically this is the ideal of functions on $V(g)$ which vanish on $V(f) \cap V(g)$, which is nonzero because $V(f)$ does not contain $V(g)$. Let $\mathfrak{p}$ be an associated point of the $A$-module $I$.

Since $I$ is an ideal in $A/gA$, we see that $\mathfrak{p}$ is an associated point of the Cartier divisor $V(g)$. It remains to show that $\tfrac{f}{g} \notin A_{\mathfrak{p}}$. Suppose $\tfrac{f}{g} = \tfrac{f'}{g'}$ in $A_{\mathfrak{p}}$, so that there exists $s \in A \setminus \mathfrak{p}$ with $sfg' = sf'g$. If $g' \notin \mathfrak{p}$, then the equation $sfg' = sf'g$ implies that $f$ vanishes in $(A/gA)_{\mathfrak{p}}$, so that $I_{\mathfrak{p}} = 0$. But $\mathfrak{p}$ is an associated point of $I$, which means it is an irreducible component of the support of $I$. This is a contradiction, so in fact $g' \in \mathfrak{p}$.

$\Box$

Proof of Proposition 2. First we prove that normality is equivalent to condition 1 under the given assumptions.

For the “only if” direction, observe that since all the one-dimensional local rings of a noetherian normal scheme are integrally closed noetherian domains, they are discrete valuation rings (see e.g. Proposition 9.2 in Atiyah-Macdonald), and in particular regular. Let $Y \subset X$ be an effective Cartier divisor and $Z \subset Y$ an associated point of $Y$. We need to show that $Z$ is an irreducible component of $Y$, or equivalently that $Z$ has codimension one in $X$. Shrinking $X$ if necessary, we can take the Cartier divisor $Y = \text{Spec } A/fA$ to be principal, cut out by some nonzero function $f \in A$. Now $Z$ corresponds to a prime $\mathfrak{p} \in \text{Spec } A$ which is the annihilator of a nonzero function on $Y$, so there is some $g \in A \setminus fA$ such that

$\mathfrak{p} = \{ a \in A \ | \ ag \in fA \}$.

We need to prove $A_{\mathfrak{p}}$ is one-dimensional, for which it suffices to show that $\mathfrak{p}A_{\mathfrak{p}}$ is principal. To this end we may localize further and replace $A$ by $A_{\mathfrak{p}}$, so that in particular $A$ is an integrally closed noetherian local domain with maximal ideal $\mathfrak{p}$. Consider the fractional ideal

$\mathfrak{p}^{-1} = \{ x \in K \ | \ x\mathfrak{p} \subset A \}$,

where $K$ is the fraction field of $A$, and form the product $\mathfrak{p}\mathfrak{p}^{-1}$. The reader may object that we do not know that $\mathfrak{p}$ is invertible, but that can be seen as follows: since $\mathfrak{p} \subset \mathfrak{p}\mathfrak{p}^{-1} \subset A$ and $\mathfrak{p}$ is maximal, we must have $\mathfrak{p}\mathfrak{p}^{-1} = \mathfrak{p}$ or $\mathfrak{p}\mathfrak{p}^{-1} = A$. In the former case the elements of $\mathfrak{p}^{-1}$ are integral over $A$ because they stabilize the finitely generated $A$-submodule $\mathfrak{p} \subset K$, so $\mathfrak{p}^{-1} = A$ by our normality assumption. But now by the definition of $\mathfrak{p}$ we have $g/f \in \mathfrak{p}^{-1} = A$, i.e. $g \in fA$, a contradiction. Thus $\mathfrak{p}$ is indeed invertible, so that there exist some $x_1,\cdots,x_n \in \mathfrak{p}$ and $y_1,\cdots,y_n \in \mathfrak{p}^{-1}$ such that $\sum x_iy_i =1$. But then $x_iy_i \notin \mathfrak{p}$ for some $i$, so that $x_i^{-1} = (x_iy_i)^{-1}y_i \in \mathfrak{p}^{-1}$ has the property that $x_i^{-1}\mathfrak{p} = A$, i.e. $\mathfrak{p} = x_iA$.

The “if” direction is easier. Lemma 1 says that $A$ is the intersection in its fraction field of the local rings $A_{\mathfrak{p}}$ where $\mathfrak{p}$ is an associated point of an effective Cartier divisor on $X$, so it suffices to show that $A_{\mathfrak{p}}$ is normal under this condition on $\mathfrak{p}$. Condition 1 tells us that $\mathfrak{p}$ has codimension one, and $X$ is regular in codimension one.

Now let us prove that condition 1 implies condition 2 for $X$ regular in codimension one. The assertion is local, so it suffices to consider $\mathscr{L} = \mathscr{O}_X$. Thus we are trying to prove that $A \to \mathscr{O}_X(U)$ is an isomorphism. The map is injective because $X$ is integral, and for surjectivity suppose we are given $f \in \mathscr{O}_X(U)$. Then $f$ is regular along every subscheme of codimension one, which includes all associated points of effective Cartier divisors on $X$ by condition 1. Thus $f \in A$ by Lemma 1.

Next we show that condition 2 implies condition 1 under our hypotheses. If $f$ is a rational function on $X$ which is regular along every subscheme of codimension one, then $f$ is regular, since condition 2 says that $f$ can be extended over loci of codimension at least two. Thus $A = \cap A_{\mathfrak{p}}$ where $\mathfrak{p}$ runs over prime ideals of height one. Since the $A_{\mathfrak{p}}$ are normal so is $A$, and we proved that normality implies condition 1.

Finally, we check that condition 3 is equivalent to condition 2. For now $\mathscr{F}$ is any coherent sheaf. Given an open subscheme $U \subset X$, write $j : U \to X$ for the inclusion, and consider the short exact sequence

$0 \to \mathscr{O}_X \to j_*\mathscr{O}_U \to (j_*\mathscr{O}_U)/\mathscr{O}_X \to 0$.

This induces the long exact sequence

$\cdots \to \text{Hom}(\mathscr{F}|_U,\mathscr{O}_U) \to \text{Hom}(\mathscr{F},(j_*\mathscr{O}_U)/\mathscr{O}_X)$

$\to \text{Ext}^1(\mathscr{F},\mathscr{O}_X) \to \text{Ext}^1(\mathscr{F}|_U,\mathscr{O}_U) \to \cdots$.

Now let $U$ be the complement of the support of $\mathscr{F}$, so we obtain an isomorphism

$\text{Hom}(\mathscr{F},(j_*\mathscr{O}_U)/\mathscr{O}_X) \cong \text{Ext}^1(\mathscr{F},\mathscr{O}_X)$.

If condition 2 holds and the support of $\mathscr{F}$ has codimension at least 2, then the left hand side vanishes, so that condition 3 holds as well. On the other hand, if we assume condition 3 and $Z = X \setminus U$ has codimension at least 2, we can take $\mathscr{F} = \mathscr{O}_Z$. The left side is identified with $\Gamma(Z,(j_*\mathscr{O}_U)/\mathscr{O}_X)$, and since $Z$ is affine the vanishing of the global sections implies $(j_*\mathscr{O}_U)/\mathscr{O}_X = 0$ as desired.

$\Box$.

Proof of Proposition 1. First assume that $X$ is locally factorial. We omit the standard argument that a unique factorization domain is normal. To prove condition 2, note that $X$ is reduced, so we may apply Corollary 7.1.19 of Qing Liu’s Algebraic Geometry and Arithmetic Curves to identify isomorphism classes of line bundles on $X$ or $U$ with the class group of Cartier divisors. Proposition 7.2.16 of loc. cit. says that the Cartier class group is identified with the Weil class group (actually, the hypotheses of that result are that $X$ is noetherian, regular, and integral; but the proof invokes our main theorem and then actually uses local factoriality). Now recall that the Weil class group is insensitive to the removal of loci of codimension at least two.

Now suppose that conditions 1 and 2 hold. Recall that a noetherian domain is a unique factorization domain if and only if every height one prime ideal is principal, so it suffices to show that the local rings of $A$ have the latter property. In fact, we will prove that every prime Weil divisor $Y \subset X$ is a Cartier divisor, which means that the ideal sheaf $\mathscr{I}_Y$ is a line bundle. Since the local ring of $X$ at $Y$ is a discrete valuation ring, the restriction of $\mathscr{I}_Y$ to the generic point of $Y$ is a trivial line bundle. Thus there is an open set $V \subset X$ with $V \cap Y$ dense in $Y$ such that $\mathscr{I}_Y$ is a line bundle over $V$. We have $\mathscr{I}_Y|_{X \setminus Y} = \mathscr{O}_{X \setminus Y}$, so that $\mathscr{I}_Y$ is a line bundle over $U = (X \setminus Y) \cup V$. Observe that the complement of $U$ has codimension at least two in $X$, so there is a line bundle $\mathscr{L}$ with $\mathscr{L}|_U \cong \mathscr{I}_Y|_U$. The map $\mathscr{L}|_U \to \mathscr{O}_U$ induces $\varphi : \mathscr{L} \to j_*\mathscr{O}_U \cong \mathscr{O}_X$, where $j : U \to X$ and the latter isomorphism uses condition 2 of Proposition 2. Since $X$ is integral $\varphi$ must be injective, and we claim that $\varphi$ maps $\mathscr{L}$ isomorphically onto $\mathscr{I}_Y$. To see that $\varphi$ lands in $\mathscr{I}_Y$, notice that $\mathscr{L} \to \mathscr{O}_Y$ factors through $\mathscr{L}|_Y \to \mathscr{O}_Y$, which vanishes because it is a generically zero map of line bundles on $Y$. The image $\varphi(\mathscr{L}) = \mathscr{I}_Z$ where $Z \subset X$ is a Cartier divisor, and we have just shown that $Y \subset Z$. Now $Y \cap U = Z \cap U$, and since $X \setminus U$ has codimension at least two this implies that $Y$ is the maximal reduced closed subscheme of $Z$, and that any component of $Z \setminus Y$ is an embedded point of $Z$. Thus $Y = Z$ by condition 1 of Proposition 2.

$\Box$

Finally we come to the proof of the theorem, which ingeniously uses the notion of determinant of a perfect complex.

If $\mathscr{E}$ is a vector bundle on a scheme $S$, recall that the determinant $\text{det } \mathscr{E}$ is the line bundle on $S$ obtained by taking the top exterior power of $\mathscr{E}$ locally (the rank of $\mathscr{E}$ may vary but is at least locally constant). More generally, if $\mathscr{E}^{\bullet}$ is a bounded complex of vector bundles on $S$, we put

$\det \mathscr{E}^{\bullet} = \bigotimes_{n \in \mathbb{Z}} (\det \mathscr{E}^n)^{(-1)^n}$.

This construction is compatible with localization and functorial in the sense that a quasi-isomorphism of perfect complexes induces an isomorphism between their determinant lines.

We can even apply $\text{det}$ to a coherent sheaf if it has a finite resolution by vector bundles. The result will not depend on the choice of resolution by the functoriality mentioned above. Any coherent sheaf on a regular quasi-projective variety has such a resolution, although we will not use this fact. We only need the following lemma.

For the rest of this post, $A$ will be a regular local ring of dimension $n$ with maximal ideal $\mathfrak{m}$ and residue field $k := A/\mathfrak{m}$.

Lemma 2 Any finitely generated $A$-module $M$ has a free resolution of length at most $n$.

Proof. We can construct a free resolution $F^{\bullet}$ of $M$, possibly of infinite length, in the following manner. Instead of picking an arbitrary set of generators of $M$, we lift a basis of $M/\mathfrak{m}M$: by Nakayama’s lemma this is the same as a minimal set of generators for $M$. This minimal set of generators yields an epimorphism $F^0 \to M$, where $F^0$ is a free module of the appropriate rank. Similarly, one can choose a minimal set of generators for the kernel of $F^0 \to M$, thereby obtaining $F^1 \to F^0$, and so on. It is not hard to see that this minimality condition is equivalent to the vanishing of all differentials in the complex $k \otimes_A F^{\bullet}$. But then we have

$k \otimes_A F^{-i} = H^{-i}(k \otimes_A F^{\bullet}) = \text{Tor}^A_i(k,M),$

so it suffices to show that $\text{Tor}^A_i(k,M) = 0$ for $i > n$, since this implies $F^{-i} = 0$ for $i > n$ (no need to apply Nakayama’s lemma: this is obvious because $F^{-i}$ is free).

For this, recall the existence of the Koszul resolution $K^{\bullet}$, which is a free resolution of the $A$-module $k$ of length $n$. Thus for $i > n$ we have

$\text{Tor}^A_i(k,M) = H^{-i}(K^{\bullet} \otimes M) = 0$.

$\Box$

Proof of the theorem. We reduce immediately to the local case, so $A$ is still a regular local ring and $X = \text{Spec } A$ as usual. First we must prove that $X$ is normal, for which it suffices to show that condition 3 of Proposition 2 holds (note that $A$ is a domain because its associated graded algebra is isomorphic to a polynomial ring, and in particular is a domain). So fix an $A$-module $M$ whose support has codimension at least two. This implies that we can find $f,g \in \mathfrak{m}$ satisfying $fM = gM = 0$ which are a regular sequence in the sense that if $I = fA + gA$, then the Koszul complex

$0 \to A \to A \oplus A \to A \to A/I \to 0$

is exact, where the first nonzero arrow sends $1 \mapsto (g,-f)$ and the second sends $(1,0) \mapsto f$, $(0,1) \mapsto g$. Consider the following segment of the long exact sequence gotten by applying $\text{Hom}(M,-)$:

$\cdots \to \text{Hom}(M,A) \to \text{Hom}(M,A/I) \to \text{Ext}^1(M,A) \to \text{Ext}^1(M,A) \to \cdots$.

Now $\text{Hom}(M,A) \to \text{Hom}(M,A/I)$ is an isomorphism because $IM = 0$, and $\text{Ext}^1(M,A) \to \text{Ext}^1(M,A)$ vanishes for the same reason. It follows that $\text{Ext}^1(M,A) = 0$ as desired.

Finally, we must prove condition 2 of Proposition 1 holds, and it is now that our investment in geometry pays off. We reduce to the following situation: $x \in X$ is the closed point, $U = X \setminus \{ x \}$, and $\mathscr{L}$ is a line bundle on $U$ which we want to extend to $X$. First, we claim we can extend $\mathscr{L}$ to a coherent sheaf on $X$. Indeed, if $j : U \to X$ is the inclusion, then $j_*\mathscr{L}$ is a quasi-coherent extension of $\mathscr{L}$, and since $X$ is noetherian $j_*\mathscr{L}$ is the filtered colimit of its coherent subsheaves. Thus there exists a coherent $\mathscr{F} \subset j_*\mathscr{L}$ such that $\mathscr{F}|_U = \mathscr{L}$. Now apply Lemma 2 to $\Gamma(X,\mathscr{F})$, so that we obtain a bounded free resolution of $\mathscr{F}$ and hence can make sense of $\text{det } \mathscr{F}$. This is a line bundle on $X$ with the property that

$(\text{det } \mathscr{F})|_U = \text{det}(\mathscr{F}|_U) = \text{det } \mathscr{L} = \mathscr{L}$.

$\Box$

Sorry for the obscenely long post, but after all, it has been almost a year.

Categories: algebraic geometry